Question

The voltage in an electrical circuit is multiplied by itself each time it is reduced. The voltage is 27/125 of a volt and it has been reduced 3 times. Write the voltage in exponential form. What was the original voltage in the circuit.

Answer 1

Answer:

The voltage in exponential form would be $(\frac{3}{5})^3$.

The original voltage in the circuit is $\frac{3}{5}$ of a volt.

Step-by-step explanation:

Let x represent the original voltage.

We have been given that the voltage in an electrical circuit is multiplied by itself each time it is reduced. The voltage is 27/125 of a volt and it has been reduced 3 times.

The original voltage multiplied by itself for 3 times would be $x\cdot x\cdot x=x^3$.

Now, we will equate $x^3$ with 27/125 as:

$x^3=\frac{27}{125}$

We know that 27 is cube of 3 and 125 is cube of 5, so we can represent our equation as:

$x^3=\frac{3^3}{5^3}$

$x^3=(\frac{3}{5})^3$

Therefore, the voltage in exponential form would be $(\frac{3}{5})^3$

Now, we will take cube root of both sides to find original voltage in the circuit as:

$\sqrt[3]{x^3} =\sqrt[3]{(\frac{3}{5})^3}$

Using property $\sqrt[n]{a^n} =a$, we will get:

$x =\frac{3}{5}$

Therefore, the original voltage in the circuit is $\frac{3}{5}$ of a volt.